HDU 1839 Delay Constrained Maximum Capacity Path(最短路+二分)

Delay Constrained Maximum Capacity Path

                                                                                Time Limit: 10000/10000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
                                                                                                     Total Submission(s): 1514    Accepted Submission(s): 480

Problem Description

Consider an undirected graph with N vertices, numbered from 1 to N, and M edges. The vertex numbered with 1 corresponds to a mine from where some precious minerals are extracted. The vertex numbered with N corresponds to a minerals processing factory. Each edge has an associated travel time (in time units) and capacity (in units of minerals). It has been decided that the minerals which are extracted from the mine will be delivered to the factory using a single path. This path should have the highest capacity possible, in order to be able to transport simultaneously as many units of minerals as possible. The capacity of a path is equal to the smallest capacity of any of its edges. However, the minerals are very sensitive and, once extracted from the mine, they will start decomposing after T time units, unless they reach the factory within this time interval. Therefore, the total travel time of the chosen path (the sum of the travel times of its edges) should be less or equal to T.

Input

The first line of input contains an integer number X, representing the number of test cases to follow. The first line of each test case contains 3 integer numbers, separated by blanks: N (2 <= N <= 10.000), M (1 <= M <= 50.000) and T (1 <= T <= 500.000). Each of the next M lines will contain four integer numbers each, separated by blanks: A, B, C and D, meaning that there is an edge between vertices A and B, having capacity C (1 <= C <= 2.000.000.000) and the travel time D (1 <= D <= 50.000). A and B are different integers between 1 and N. There will exist at most one edge between any two vertices.

Output

For each of the X test cases, in the order given in the input, print one line containing the highest capacity of a path from the mine to the factory, considering the travel time constraint. There will always exist at least one path between the mine and the factory obbeying the travel time constraint.

Sample Input

22 1 101 2 13 104 4 201 2 1000 152 4 999 61 3 100 153 4 99 4

 

Sample Output

1399

Author

Mugurel Ionut Andreica

题目大意：给出n个点，m条路，以及每条路的端点，容纳量，通过时间，问T时间内可以通过的最大容量

其实一开始看题时以为在TH迪的引导下以为是最大流，还好坚持了自我，一直坚定的按照最短路做，其实正解也是最短路SPFA 算法，敲一个标准的版+判断一下点的容纳值就好了，二分的做法着实没想到，代码如下：

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <vector>#include <map>#include <list>#include <cmath>#include <queue>#include <stack>#include <set>#include <algorithm>#define LL long long#define INF 0x3f3f3f3f#define RR freopen("in.txt","r",stdin)#define WW freopen("out.txt","w",stdout)#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1using namespace std;const int N = 10010;const int M = 500010;struct node{    int v,w,t;    int next;} edge[M*2];int head[M];bool vis[N];int cnt;int d[N];int Cap[M];int n,m,T;bool cmp(int a,int b){    return a > b;}void SPFA(int Val){    queue<int >Q;    memset(vis,false,sizeof(vis));    memset(d, INF, sizeof(d));    Q.push(1);    d[1] = 0;    while(!Q.empty())    {        int u = Q.front();        Q.pop();        vis[u] = false;        for(int i = head[u]; i != -1; i= edge[i].next)        {            int v = edge[i].v;            int w = edge[i].w;            int t = edge[i].t;            if(Val <= w)            {                if(d[v] > d[u] + t)                {                    d[v] = d[u] + t;                    if(!vis[v])                    {                        vis[v] = true;                        Q.push(v);                    }                }            }        }    }}void Add(int u,int v,int w,int t){    edge[cnt].v = v;    edge[cnt].w = w;    edge[cnt].t = t;    edge[cnt].next = head[u];    head[u] = cnt++;}void Solve(int l,int r){    if(l >= r)    {        printf("%d\n",Cap[l]);        return ;    }    int mid = (l + r) >> 1;    SPFA(Cap[mid]);    if(d[n] == INF || d[n] > T)        Solve(mid+1, r);    else        Solve(l,mid);}int main(){    int X;    scanf("%d",&X);    while(X--)    {        cnt = 0;        scanf("%d%d%d",&n,&m,&T);        memset(head, -1, sizeof(head));        for(int i=0; i<m; i++)        {            int u,v,w,t;            scanf("%d%d%d%d",&u,&v,&w,&t);            Add(u,v,w,t);            Add(v,u,w,t);            Cap[i] = w;        }        sort(Cap,Cap+m,cmp);        Solve(0, m-1);    }    return 0;}